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Derivative

lf a+bxey/x...

Question

lf $(a + b x) e^{y / x} = x$ then $x^{3} y_{2} =$

A

x y_{1} - y

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B

(y_{1} - x y)^{2}

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C

(x y_{1} - y)^{2}

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D

(y_{1} - x y)^{3}

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Solution

The correct option is D $(x y_{1} - y)^{2}$
$y = x (l o g x - l o g (a + b x))$
$y^{^{'}} = (l o g x - l o g (a + b x)) + x (\frac{1}{x} - \frac{b}{a + b x})$

$y^{^{''}} = \frac{1}{x} - \frac{b}{a + b x} - \frac{b a}{(a + b x)^{2}}$

$x^{3} y^{^{''}} = \frac{a^{2} x^{2}}{(a + b x)^{2}}$

$x^{3} y^{^{''}} = (x y_{1} - y)^{2}$

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